Zeuner Valve Diagram.
bringing them nearer together ; the reverse happening when the
governors fall, and thus is obtained automatically an early or late
cut-off respectively, A dash-pot within P damps small vibrations
on the governor, entrance or exit of air being adjusted by screw Q.
The steam valves B B, being double-beat, are balanced, besides
requiring only half the lift of a single valve.
Zeuner's Valve Diagram is a graphic and ready means
of finding the various positions 'of the engine crank where
admission, cut-off, release, compression, &c., take place with a D
slide valve, when the valve dimensions are known; or, conversely,
of finding valve dimensions when certain crank positions are
given. . •:•'..
Imagine a valve without lap, and let CD he the eccentric
throw or radius at (i) Fig. 653, When the eccentric is at D, the
valve is closed, and, when moved to H, the opening to steam on
the left is CG. Turn G round to LJ then, C.L is steam opening for
eccentric position H. A series of points such as L may be found
and the curve c L B drawn, whose radii vector' shew gradual
opening and closing of the left-hand steam port, The left diagram
being obtained similarly, join A K, Then triangles c A K, cj F are
similar and equal, and A K c is a right angle ', the two polar curves
are circles, and while circle c B shews steam opening, circle c A
represents the opening to exhaust', together being known as curves
of position for a valve without lap.
Taking a valve having lap, both to steam and exhaust, its
position curves are those at (2) Fig. 653. For the opening either
to steam or exhaust will be that at (i) less the respective lap.
At centre c, strike arc M N with radius =? lap, while p P = exhaust
lap. Then 4 B is full opening to steam at left-hand port, eccentric
being at B ; Q is admission position, and R that of cut-off. Simi-
larly AS is full opening to exhaust at right-hand port, eccentric
being at A ; T the release, and s the compression position.
We must now translate eccentric position into crank position.
Still assuming .a right-handed rotation, we, must turn back in a
left-handed direction all the eccentric positions, through the angle
by which the eccentric leads the crank, to effect the above purpose.
This angle is 90° plus the angle of advance.* The change has
* Angle pf advance == the a$g!e whose sine = (l^p + lead) -f throw,